A stone is thrown horizontally from the top of a bridge 19.6 m above the street below. The initial velocity of the stone is 5 m/sec  when it hits the floor of street, direction of the velocity of projectile is: (g=9.8 m/s2)

given height from where stone is projected=h=19.6m; acceleration due to gravity =9.8m/s2  initial velocity of stone=u=ux=5m/s;  here vertical component of initial velocity=uy=0 Vy2=uy2+2gh vy=2gh=2×9.8×19.6=19.6m/s  Tanθ=VyVx=19.65=3.92 ⇒θ=tan-1(3.92) ⇒θ=760